Chapter 5 Thermochemistry

Topics  Energy and energy changes  Introduction to thermodynamics  Enthalpy  Calorimetry  Hess’s Law  Standard enthalpies of formation

Copyright McGraw-Hill Energy and Energy Changes  Energy is involved in all types of physical and chemical changes  Energy: the capacity to do work or transfer heat  All forms of energy are either Kinetic Kinetic Potential Potential

 Kinetic energy - energy of motion Defining equation Defining equation Where m is mass and u is velocity Where m is mass and u is velocity Thermal energy - one form of kinetic energy associated with random motion Thermal energy - one form of kinetic energy associated with random motion Changes in thermal energy are monitored via changes in temperatureChanges in thermal energy are monitored via changes in temperature

 Potential energy - energy of position or composition Chemical energy is stored within structural units of chemical substances. Chemical energy is stored within structural units of chemical substances. Electrostatic energy is energy resulting from the interaction of charged particles. Electrostatic energy is energy resulting from the interaction of charged particles. Dependent on charges and distance between charges (Q = charge and d = distance )Dependent on charges and distance between charges (Q = charge and d = distance ) Defining equationDefining equation + E el : repulsive+ E el : repulsive − E el : attractive − E el : attractive

 Law of conservation of energy  Energy can be converted from one form to another but not created or destroyed. The total amount of energy in the universe is constant. The total amount of energy in the universe is constant. Example Example A chemical reaction (potential) gives off heat (thermal)A chemical reaction (potential) gives off heat (thermal)

System is the part of the universe of interest. System is the part of the universe of interest. ExampleExample The reactants NaOH and HCl The reactants NaOH and HCl Surroundings are the rest of the universe. Surroundings are the rest of the universe. ExampleExample When heat is given off from the reaction of NaOH and HCl, the energy is transferred from the system to the surroundings. When heat is given off from the reaction of NaOH and HCl, the energy is transferred from the system to the surroundings. system: what is inside the container. Reactants or products of a chemical reaction system: what is inside the container. Reactants or products of a chemical reaction surroundings: container,room, etc. surroundings: container,room, etc. Energy changes in chemical reactions

Copyright McGraw-Hill The study of the transfer of heat (thermal energy) in chemical reactions. The study of the transfer of heat (thermal energy) in chemical reactions. Exothermic - transfer of heat from the system to the surroundings Exothermic - transfer of heat from the system to the surroundings 2H 2 (g) + O 2 (g)  2H 2 O(l) + energy 2H 2 (g) + O 2 (g)  2H 2 O(l) + energy Endothermic - the transfer of heat from the surroundings to the system Endothermic - the transfer of heat from the surroundings to the system energy + 2HgO(s)  2Hg(l) + O 2 (g) energy + 2HgO(s)  2Hg(l) + O 2 (g) Thermochemistry

Transfer of Energy Exothermic process energy is produced in reaction energy is produced in reaction energy flows out of system energy flows out of system container becomes hot when touched container becomes hot when touched CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g) + heat CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g) + heat  Endothermic process energy is consumed by the reaction energy is consumed by the reaction energy flows into the system energy flows into the system container becomes cold when touched container becomes cold when touched N 2 (g) + O 2 (g) + Energy 2NO(g) N 2 (g) + O 2 (g) + Energy 2NO(g)

Comparison of Endothermic and Exothermic Processes

Joule (J) is the SI unit for energy. Joule (J) is the SI unit for energy. The amount of energy possessed by a 2 kg mass moving at a speed of 1 m/sThe amount of energy possessed by a 2 kg mass moving at a speed of 1 m/s Units of Energy

Copyright McGraw-Hill Calorie (cal) - commonly used on food labels Calorie (cal) - commonly used on food labels 1 cal J 1 cal J 1000 cal = 1 Cal = 1 kcal 1000 cal = 1 Cal = 1 kcal Food calories (Cal) are really 1000 calories (cal). Food calories (Cal) are really 1000 calories (cal).

Calculate the kinetic energy of a neon atom moving at a speed of 98 m/s. Exampl e

5.2 Introduction to Thermodynamics  Types of systems: open (exchange of mass and energy) open (exchange of mass and energy) closed (exchange of energy) closed (exchange of energy) isolated (no exchange) isolated (no exchange)

 State functions depend only on initial and final states of the system and not on how the change was carried out. Energy (E) Energy (E) Pressure (P) Pressure (P) Volume (V) Volume (V) Temperature (T) Temperature (T) State function

State Function  energy change is independent of pathway thus it is a state function (state property) (state property)  State function: a property of a system that depends only on its present state (it does not depend on system ’ s past or future, i.e, it does not depend on how the system arrived at the present state)  work and heat depend on pathway so they  work and heat depend on pathway so they are not state functions

First Law of Thermodynamics  Energy can be converted from one form to another but cannot be created or destroyed. Based on the law of conservation of energy Based on the law of conservation of energy  Internal energy (U) It is the sum of potential and kinetic energy of all particles in the system  U = P E + K E Kinetic energy - molecular motion Kinetic energy - molecular motion Potential energy - attractive/repulsive interactions Potential energy - attractive/repulsive interactions

 The change in internal energy of a system between final (f) and initial (i) states is defined as:  U = U f  U i  U = U f  U i  For a chemical system Cannot calculate the total internal energy with any certainty Cannot calculate the total internal energy with any certainty Can calculate the change in energy of the system experimentally Can calculate the change in energy of the system experimentally  U = U (products)  U (reactants)  U = U (products)  U (reactants)

Consider: Consider: S(s) + O 2 (g)  SO 2 (g) S(s) + O 2 (g)  SO 2 (g) This reaction releases heat, therefore  U is negative. This reaction releases heat, therefore  U is negative.

 U system +  U surroundings = 0  When a system releases heat, some of the chemical energy is released as thermal energy to the surroundings but this does not change the total energy of the universe.

 U system =   U surroundings  When a system undergoes a change in energy, the surroundings must undergo a change in energy equal in magnitude and opposite in sign.

 where q is heat, w is work Work and heat  U can be changed (- or + ∆U) by a flow of work, heat, or both  U sys = q + w  U sys = q + w

Sign Conventions of q and w

Calculate the overall change in internal energy for a system that absorbs 125 J of heat and does 141 J of work on the surroundings. q is + (heat absorbed) w is  (work done)  U sys = q + w = (+125 J) + (  141J) =  16 J

Work  common types of work Expansion- work done by the system Expansion- work done by the system Compression- work done on the system Compression- work done on the systemexpansion+∆V-wcompression-∆V+w P is external pressure Work is force applied over distance

Pressure-volume work, w, done by a system is Pressure-volume work, w, done by a system is w =  P  V w =  P  V  U sys = q + w  U sys = q + w  U sys = q  P  V  U sys = q  P  V Constant volume,  V = 0 Constant volume,  V = 0 q v =  U sys q v =  U sys q v means a constant-volume process q v means a constant-volume process 5.3 Reactions carried out at constant volume or at constant pressure or at constant pressure

  U = q + w  U = q p  P  V q p =  U + P  V q p =  U + P  V Reactions carried out at constant pressure

Enthalpy and enthalpy changes Enthalpy and enthalpy changes  Enthalpy is a property of a system  definition: H = U + PV  since U, P and V are all state functions, then H is also a state function  The term enthalpy is composed of the prefix en-, meaning "to put into" and the Greek word - thalpein, meaning "to heat", that is  The term enthalpy is composed of the prefix en-, meaning "to put into" and the Greek word - thalpein, meaning "to heat", that is “to put heat into”Greek

Enthalpy change Enthalpy change H = U + PV The change in enthalpy  H =  U +  (PV) When pressure is held constant  H =  U + P  V Since q p =  U + P  V q p =  H

At constant pressure At constant pressure q p =  H Enthalpy of reaction Enthalpy of reaction   H is + for endothermic changes.   H is − for exothermic changes.

Copyright McGraw-Hill 2009 Equations that represent both mass and enthalpy changes Equations that represent both mass and enthalpy changes  Endothermic reaction H 2 O(s)  H 2 O(l)  H = kJ/mol  Exothermic reaction Thermochemical Equations

Copyright McGraw-Hill Comparison of Endothermic and Exothermic Changes

Always specify state of reactants and products. Always specify state of reactants and products. When multiplying an equation by a factor (n), multiply the  H value by same factor. When multiplying an equation by a factor (n), multiply the  H value by same factor. Reversing an equation changes the sign but not the magnitude of the  H. Reversing an equation changes the sign but not the magnitude of the  H. Thermochemical equation guidelines

Copyright McGraw-Hill Given the following equation, C 6 H 12 O 6(s) + 6O 2(g) → 6CO 2(g) + 6H 2 O (l) C 6 H 12 O 6(s) + 6O 2(g) → 6CO 2(g) + 6H 2 O (l) Δ H =  2803 kJ/mol Δ H =  2803 kJ/mol calculate the energy released when g of glucose is burned in oxygen.

5.4 Calorimetry  The science of measuring heat changes  calorimeter- device used to experimentally find the heat associated with a chemical reaction  substances respond differently when heated ( to raise T for two substances by 1 degree, they require different amount of heat)

Specific heat (s)  Specific heat (s) - the amount of heat required to raise the temp of 1 g of a substance by 1  C. Units: J/g  C Units: J/g  C Relation to amount of heat (q) Relation to amount of heat (q) where q is heat, m is mass, s is specific heat  and  T = change in temp (  T = T final  T initial ) (  T = T final  T initial )

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Copyright McGraw-Hill  Heat capacity (C) - the amount of heat required to raise the temp of an object by 1  C. Units: J/  C Units: J/  C Relation to amount of heat (q) Relation to amount of heat (q) where q is heat, C is heat capacity where q is heat, C is heat capacity  and  T = change in temp (  T = T final  T initial ) ‏

Calculate the amount of energy required to heat 95.0 grams of water from 22.5  C to 95.5  C.   T = T final – T initial = 95.5 o C − 22.5 o C  T = 73.0 o C q = (95.0 g) (4.184 J/g  C) (73.0  C) ‏ q = 2.90 x 10 4 J or 29.0 kJ q = 2.90 x 10 4 J or 29.0 kJ

Constant-Pressure Calorimetry  uses simplest calorimeter (like coffee-cup calorimeter) since it is open to air  used to find heat exchange between the system and between the system and surroundings of a chemical surroundings of a chemical reaction reaction i.e., changes in enthalpy i.e., changes in enthalpy for reactions occurring in a solution since q P = ∆H since q P = ∆H System: Reactants & Products Surroundings: the water

System: reactants and products Surroundings: water Surroundings: water Assuming that the calorimeter Assuming that the calorimeter does not leak or absorb heat does not leak or absorb heat coffee cup calorimeter Constant-pressure calorimetry Constant-pressure calorimetry For an exothermic reaction:

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Copyright McGraw-Hill A metal pellet with a mass of grams at an original temperature of 92.5  C is dropped into a calorimeter\ with grams of water at an original temperature of 23.1  C. The final temperature of the water and the pellet Is 26.8  C. Calculate the heat capacity and the specific heat for the metal. Example

q water = ms  T q water = ms  T = ( g) (4.184 J/g  C) (3.7  C) = ( g) (4.184 J/g  C) (3.7  C) = 2300 J (water gained energy) = 2300 J (water gained energy) = J (pellet released energy) = J (pellet released energy) Heat capacity of pellet: q = C  T Heat capacity of pellet: q = C  T C = q/  T C = q/  T =  2300 J/  65.7  C = 35 J/  C =  2300 J/  65.7  C = 35 J/  C Specific heat of pellet: J/g o C Specific heat of pellet: J/g o C

bomb calorimeter Constant-volume calorimetry Isolated system  Here, the ∆V = 0 so -P∆V = w = 0 ∆U = q + w = q V for constant volume

Typical procedure used in a bomb calorimeter Typical procedure used in a bomb calorimeter Known amount of sample placed in steel container and then filled with oxygen gasKnown amount of sample placed in steel container and then filled with oxygen gas Steel chamber submerged in known amount of waterSteel chamber submerged in known amount of water Sample ignited electricallySample ignited electrically Temperature increase of water is determinedTemperature increase of water is determined

Copyright McGraw-Hill A snack chip with a mass of 2.36 g was burned in a bomb calorimeter. The heat capacity of the calorimeter kJ/  C. During the combustion the water temp rose by 2.70  C. Calculate the energy in kJ/g for the chip

Copyright McGraw-Hill q rxn = − C cal  T q rxn = − C cal  T = − (38.57 kJ/  C) (2.70  C) = − (38.57 kJ/  C) (2.70  C) = − 104 kJ = − 104 kJ Energy content is a positive quantity. Energy content is a positive quantity. = 104 kJ/2.36 g = 104 kJ/2.36 g = 44.1 kJ/g = 44.1 kJ/g Food Calories: 10.5 Cal/g Food Calories: 10.5 Cal/g

Example 3  Compare the energy released in the combustion of H 2 and CH 4 carried out in a bomb calorimeter with a heat capacity of 11.3 kJ/°C. The combustion of 1.50 g of methane produced a T change of 7.3°C while the combustion of 1.15 g of hydrogen produced a T change of 14.3°C. Find the energy of combustion per gram for each.

Example 3  methane: CH 4  hydrogen: H 2  The energy released by H 2 is about 2.5 times the energy released by CH 4

5.5 Hess’s Law Hess’s Law: The change in enthalpy that occurs when reactants are converted to products is the same whether the reaction occurs in one step or a series of steps. It is sed for calculating enthalpy for a reaction that cannot be determined directly.

Example 1  N 2 (g) + 2O 2 (g)  2NO 2 (g)∆H = 68 kJ OR  N 2 (g) + O 2 (g)  2NO(g)∆H = 180 kJ 2NO(g) + O 2 (g)  2NO 2 (g)∆H = -112 kJ N 2 (g) + 2O 2 (g)  2NO 2 (g)∆H = 68 kJ

Rules 1. If a reaction is reversed, the sign of ∆H must be reversed as well. because the sign tells us the direction of heat flow at constant P because the sign tells us the direction of heat flow at constant P 2. The magnitude of ∆H is directly proportional to quantities of reactants and products in reaction. If coefficients are multiplied by an integer, the ∆H must be multiplied in the same way. because ∆H is an extensive property because ∆H is an extensive property

Copyright McGraw-Hill Given the following equations: H 3 BO 3 (aq)  HBO 2 (aq) + H 2 O(l)  H rxn =  0.02 kJ H 2 B 4 O 7 (aq) + H 2 O(l)  4 HBO 2 (aq)  H rxn =  11.3 kJ H 2 B 4 O 7 (aq)  2 B 2 O 3 (s) + H 2 O(l)  H rxn = 17.5 kJ Find the  H for this overall reaction. 2H 3 BO 3 (aq)  B 2 O 3 (s) + 3H 2 O(l)

2H 3 BO 3(aq)  2HBO 2(aq) + 2H 2 O (l) x 2  H rxn = 2( − 0.02 kJ) = − 0.04 kJ 2HBO 2(aq)  1/2H 2 B 4 O 7(aq) + 1/2H 2 O (l) reverse, ÷2  H rxn = kJ/2 = 5.65 kJ  H rxn = kJ/2 = 5.65 kJ 1/2H 2 B 4 O 7(aq)  B 2 O 3(s) + 1/2H 2 O (l) ÷ 2  H rxn = 17.5 kJ/2 = 8.75 kJ 2H 3 BO 3 (aq)  B 2 O 3 (s) + 3H 2 O(l)  H rxn = kJ

Copyright McGraw-Hill Standard Enthalpies of Formation Symbol:  H f  Symbol:  H f  The enthalpy change that results when 1 mole of a compound is formed from its elements in their standard states. The enthalpy change that results when 1 mole of a compound is formed from its elements in their standard states.  H f  for an element in its standard state is defined as zero. Standard state: 1 atm, 25  C Standard state: 1 atm, 25  C Values found in reference tables Values found in reference tables Used to calculate the  H  rxn Used to calculate the  H  rxn

Standard States  For a :  For a compound: for gas: P = 1 atm for gas: P = 1 atm For pure substances, it is a pure liquid or pure solid state For pure substances, it is a pure liquid or pure solid state in solution: concentration is 1 M in solution: concentration is 1 M  For an :  For an element: form that exists in at 1 atm and 25°C form that exists in at 1 atm and 25°C O: O 2 (g)K: K(s)Br: Br 2 (l)

Copyright McGraw-Hill   H  rxn =  n  H f  (products)   m  H f  (reactants) where  denotes summation where  denotes summation n and m are the coefficients in the n and m are the coefficients in the balanced equation balanced equation Defining equation for enthalpy of reaction:

Calculate the  H  rxn for the following reaction from the table of standard values. CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(l) CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(l)  H rxn  =  n  H f  (products) -  m  H f  (reactants) = [1(  393.5) + 2(  285.8)]  [1(  74.8) + 2(0)] = [1(  393.5) + 2(  285.8)]  [1(  74.8) + 2(0)] =  kJ/mol (exothermic) =  kJ/mol (exothermic) Example:

Copyright McGraw-Hill Key Points  First law of thermodynamics  Enthalpy (heat of formation; heat of reaction)  State function  Calorimetry  Specific heat  Hess’s law  Calculations involving enthalpy, specific heat, energy